# Chapter 4 – Principle of Mathematical Questions and Answers: NCERT Solutions for Class 11 Maths

Class 11 Maths NCERT book solutions for Chapter 4 - Principle of Mathematical Questions and Answers

Education Blogs Chapter 4 – Principle of Mathematical Questions and Answers: NCERT Solutions for...

Class 11 Maths NCERT book solutions for Chapter 4 - Principle of Mathematical Questions and Answers

1+3+32+…..+3n-1=(3n-1)21+3+32+…..+3n-1=(3n-1)2

P(n):1+3+32+…..+3n-1=(3n-1)2P(n):1+3+32+…..+3n-1=(3n-1)2

For n=1n=1,

L.H.S.=31−1=1L.H.S.=31−1=1

R.H.S.=(31-1)2=3-12=22=1R.H.S.=(31-1)2=3-12=22=1

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1+3+32+…..+3k-1=(3k-1)2…..(i)1+3+32+…..+3k-1=(3k-1)2…..(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

1+3+32+…..+3k-1+3(k+1)-11+3+32+…..+3k-1+3(k+1)-1

=(1+3+32+….+3k-1)+3k=(1+3+32+….+3k-1)+3k

=(3k-1)2+3k=(3k-1)2+3k Using(i)Using(i)

=(3k-1)+2.3k2=(3k-1)+2.3k2

=(1+2)3k−12=(1+2)3k−12

=3.3k−12=3.3k−12

=3k+1−12=3k+1−12

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

13+23+33+…..+n3=(n(n+1)2)213+23+33+…..+n3=(n(n+1)2)2

P(1):13+23+33+…..+n3=(n(n+1)2)2P(1):13+23+33+…..+n3=(n(n+1)2)2

L.H.S.=13=1L.H.S.=13=1

R.H.S.=(1(1+1)2)2=(1)2=1R.H.S.=(1(1+1)2)2=(1)2=1

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

13+23+33+…..+k3=(k(k+1)2)2…..(i)13+23+33+…..+k3=(k(k+1)2)2…..(i)

Consider

13+23+33+…..+k3+(k+1)313+23+33+…..+k3+(k+1)3

=(k(k+1)2)2+(k+1)3=(k(k+1)2)2+(k+1)3 Using(i)Using(i)

=k2(k+1)24+(k+1)3=k2(k+1)24+(k+1)3

=k2(k+1)2+4(k+1)34=k2(k+1)2+4(k+1)34

=(k+1)2{k2+4(k+1)}4=(k+1)2{k2+4(k+1)}4

=(k+1)2{k2+4k+4}4=(k+1)2{k2+4k+4}4

=(k+1)2(k+2)24=(k+1)2(k+2)24

=(k+1)2(k+1+1)24=(k+1)2(k+1+1)24

=((k+1)(k+1+1)2)2=((k+1)(k+1+1)2)2

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…n)=2n(n+1)1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…n)=2n(n+1)

P(n):1+11+2+11+2+3+…..+11+2+3+…n=2nn+1P(n):1+11+2+11+2+3+…..+11+2+3+…n=2nn+1

For n=1n=1,

L.H.S.=11=1L.H.S.=11=1

R.H.S.=2⋅11+1=1R.H.S.=2⋅11+1=1

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…+k)=2k(k+1)……(i)1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…+k)=2k(k+1)……(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

1+11+2+11+2+3+….+11+2+3+…+k+11+2+3+…+k+(k+1)1+11+2+11+2+3+….+11+2+3+…+k+11+2+3+…+k+(k+1)

=(1+11+2+11+2+3+…+11+2+3+…+k)+11+2+3+…+k+(k+1)=(1+11+2+11+2+3+…+11+2+3+…+k)+11+2+3+…+k+(k+1)

=2kk+1+11+2+3+…+k+(k+1)=2kk+1+11+2+3+…+k+(k+1) Using(i)Using(i)

=2kk+1+1((k+1)(k+1+1)2)=2kk+1+1((k+1)(k+1+1)2) [1+2+3…+n=n(n+1)2][1+2+3…+n=n(n+1)2]

=2kk+1+2(k+1)(k+2)=2kk+1+2(k+1)(k+2)

=2(k+1)(k+1k+2)=2(k+1)(k+1k+2)

=2(k+1)(k2+2k+1k+2)=2(k+1)(k2+2k+1k+2)

=2(k+1)[(k+1)2k+2]=2(k+1)[(k+1)2k+2]

=2(k+1)(k+2)=2(k+1)(k+2)

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)41.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4

P(n):1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4P(n):1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4

For n=1n=1,

L.H.S.=1.2.3=6L.H.S.=1.2.3=6

R.H.S.=1(1+1)(1+2)(1+3)4=1.2.3.44=6R.H.S.=1(1+1)(1+2)(1+3)4=1.2.3.44=6

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1.2.3+2.3.4+…+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4…(i)1.2.3+2.3.4+…+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

1.2.3+2.3.4+…+k(k+1)(k+2)+(k+1)(k+2)(k+3)1.2.3+2.3.4+…+k(k+1)(k+2)+(k+1)(k+2)(k+3)

={1.2.3+2.3.4+…+k(k+1)(k+2)}+(k+1)(k+2)+(k+3)={1.2.3+2.3.4+…+k(k+1)(k+2)}+(k+1)(k+2)+(k+3)

=k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3)=k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3) Using(i)Using(i)

=(k+1)(k+2)(k+3)(k4+1)=(k+1)(k+2)(k+3)(k4+1)

=(k+1)(k+2)(k+3)(k+4)4=(k+1)(k+2)(k+3)(k+4)4

=(k+1)(k+1+1)(k+1+2)(k+1+3)4=(k+1)(k+1+1)(k+1+2)(k+1+3)4

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

n∈Nn∈N :

1.3+2.32+3.33+…+n.3n=(2n-1)3n+1+341.3+2.32+3.33+…+n.3n=(2n-1)3n+1+34

P(n):1.3+2.32+3.33+…+n3n=(2n-1)3n+1+34P(n):1.3+2.32+3.33+…+n3n=(2n-1)3n+1+34

For n=1n=1,

L.H.S.=1.31=3L.H.S.=1.31=3

R.H.S.=(2.1-1)31+1+34=32+34=124=3R.H.S.=(2.1-1)31+1+34=32+34=124=3

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1.3+2.32+3.33+…+k3k=(2k-1)3k+1+34…(i)1.3+2.32+3.33+…+k3k=(2k-1)3k+1+34…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

1.3+2.32+3.33+…+k.3k+(k+1).3k+11.3+2.32+3.33+…+k.3k+(k+1).3k+1

=(1.3+2.32+3.33+…+k.3k)+(k+1).3k+1=(1.3+2.32+3.33+…+k.3k)+(k+1).3k+1

=(2k-1)3k+1+34+(k+1)3k-1=(2k-1)3k+1+34+(k+1)3k-1 Using(i)Using(i)

=(2k-1)3k+1+3+4(k+1)3k+14=(2k-1)3k+1+3+4(k+1)3k+14

=3k+1{2k-1+4(k+1)}+34=3k+1{2k-1+4(k+1)}+34

=3k+1{6k+3}+34=3k+1{6k+3}+34

=3k+1.3{2k+1}+34=3k+1.3{2k+1}+34

=3(k+1)+1{2k+1}+34=3(k+1)+1{2k+1}+34

={2(k+1)-1}3(k+1)+1+34={2(k+1)-1}3(k+1)+1+34

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]

P(n):1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]P(n):1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]

For n=1n=1,

L.H.S.=1.2=2L.H.S.=1.2=2

R.H.S.=1(1+1)(1+2)3=1.2.33=2R.H.S.=1(1+1)(1+2)3=1.2.33=2

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1.2+2.3+3.4+…+k.(k+1)=[k(k+1)(k+2)3]…(i)1.2+2.3+3.4+…+k.(k+1)=[k(k+1)(k+2)3]…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

1.2+2.3+3.4+…+k.(k+1)+(k+1).(k+2)1.2+2.3+3.4+…+k.(k+1)+(k+1).(k+2)

=[1.2+2.3+3.4+…+k.(k+1)]+(k+1).(k+2)=[1.2+2.3+3.4+…+k.(k+1)]+(k+1).(k+2)

=k(k+1)(k+2)3+(k+1)(k+2)=k(k+1)(k+2)3+(k+1)(k+2) Using(i)Using(i)

=(k+1)(k+2)(k3+1)=(k+1)(k+2)(k3+1)

=(k+1)(k+2)(k+3)3=(k+1)(k+2)(k+3)3

=(k+1)(k+1+1)(k+1+2)3=(k+1)(k+1+1)(k+1+2)3

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)31.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3

P(n):1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3P(n):1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3

For n=1n=1,

L.H.S.=1.3=3L.H.S.=1.3=3

R.H.S.=1(4.12+6.1-1)3=4+6-13=93=3R.H.S.=1(4.12+6.1-1)3=4+6-13=93=3

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1.3+3.5+5.7+…+(2k-1)(2k+1)=k(4k2+6k-1)3…(i)1.3+3.5+5.7+…+(2k-1)(2k+1)=k(4k2+6k-1)3…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

(1.3+3.5+5.7+…+(2k-1)(2k+1))+{(k+1)-1}{2(k+1)+1}(1.3+3.5+5.7+…+(2k-1)(2k+1))+{(k+1)-1}{2(k+1)+1}

=k(4k2+6k-1)3+(2k+2-1)(2k+2+1)=k(4k2+6k-1)3+(2k+2-1)(2k+2+1) Using(i)Using(i)

=k(4k2+6k-1)3+(2k+1)(2k+3)=k(4k2+6k-1)3+(2k+1)(2k+3)

=k(4k2+6k-1)3+(4k2+8k+3)=k(4k2+6k-1)3+(4k2+8k+3)

=k(4k2+6k-1)+3(4k2+8k+3)3=k(4k2+6k-1)+3(4k2+8k+3)3

=4k3+6k2-k+12k2+24k+93=4k3+6k2-k+12k2+24k+93

=4k3+18k2+23k+93=4k3+18k2+23k+93

=4k3+14k2+9k+4k2+14k+93=4k3+14k2+9k+4k2+14k+93

=k(4k2+14k+9)+1(4k2+14k+9)3=k(4k2+14k+9)+1(4k2+14k+9)3

=(k+1)(4k2+14k+9)3=(k+1)(4k2+14k+9)3

=(k+1){4k2+8k+4+6k+6-1}3=(k+1){4k2+8k+4+6k+6-1}3

=(k+1){4(k2+2k+1)+6(k+1)-1}3=(k+1){4(k2+2k+1)+6(k+1)-1}3

=(k+1){4(k+1)2+6(k+1)-1}3=(k+1){4(k+1)2+6(k+1)-1}3

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

1.2+2.22+3.22+….+n.2n=(n-1)2n+1+21.2+2.22+3.22+….+n.2n=(n-1)2n+1+2

P(n):1.2+2.22+3.22+….+n.2n=(n-1)2n+1+2P(n):1.2+2.22+3.22+….+n.2n=(n-1)2n+1+2

For n=1n=1,

L.H.S.=1.2=2L.H.S.=1.2=2

R.H.S.=(1-1)21+1+2=0+2=2R.H.S.=(1-1)21+1+2=0+2=2

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1.2+2.22+3.22+…+k.2k=(k-1)2k+1+2…(i)1.2+2.22+3.22+…+k.2k=(k-1)2k+1+2…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

{1.2+2.22+3.22+…+k.2k}+(k+1).2k+1{1.2+2.22+3.22+…+k.2k}+(k+1).2k+1

=(k-1)2k+1+2+(k+1)2k+1=(k-1)2k+1+2+(k+1)2k+1

=2k+1{(k-1)+(k+1)}+2=2k+1{(k-1)+(k+1)}+2

=2k+1.2k+2=2k+1.2k+2

=k.2(k+1)+1+2=k.2(k+1)+1+2

={(k+1)-1}2(k+1)+1+2={(k+1)-1}2(k+1)+1+2

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

12+14+18+…+12n=1-12n12+14+18+…+12n=1-12n

P(n):12+14+18+…+12n=1-12nP(n):12+14+18+…+12n=1-12n

For n=1n=1,

L.H.S.=12L.H.S.=12

R.H.S.=1-121=12R.H.S.=1-121=12

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

12+14+18+…+12k=1-12k…(i)12+14+18+…+12k=1-12k…(i)

Consider

(12+14+18+…+12k)+12k+1(12+14+18+…+12k)+12k+1

=(1-12k)+12k+1=(1-12k)+12k+1 Using(i)Using(i)

1-12k(1-12)1-12k(1-12)

=1-12k(12)=1-12k(12)

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)

P(n):12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)P(n):12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)

For n=1n=1,

L.H.S.=12.5=110L.H.S.=12.5=110

R.H.S.=16.1+4=110R.H.S.=16.1+4=110

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

12.5+15.8+18.11+…+1(3k+1)(3k+2)=k(6k+4)…(i)12.5+15.8+18.11+…+1(3k+1)(3k+2)=k(6k+4)…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

12.5+15.8+18.11+…+1(3k+1)(3k+2)+1{3(k+1)-1}{3(k+1)+2}12.5+15.8+18.11+…+1(3k+1)(3k+2)+1{3(k+1)-1}{3(k+1)+2}

=k6k+4+1(3k+3-1)(3k+3+2)=k6k+4+1(3k+3-1)(3k+3+2) Using(i)Using(i)

=k6k+4+1(3k+2)(3k+5)=k6k+4+1(3k+2)(3k+5)

=k2(3k+2)+1(3k+2)(3k+5)=k2(3k+2)+1(3k+2)(3k+5)

=1(3k+2)(k2+13k+5)=1(3k+2)(k2+13k+5)

=1(3k+2)((3k+2)(k+1)2(3k+5))=1(3k+2)((3k+2)(k+1)2(3k+5))

=(k+1)6k+10=(k+1)6k+10

=(k+1)6(k+1)+4=(k+1)6(k+1)+4

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle

of mathematical induction.

11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

P(n):11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)P(n):11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

For n=1n=1,

L.H.S.=11.2.3=16L.H.S.=11.2.3=16

R.H.S.=1.(1+3)4(1+1)(1+2)=1.44.2.3=16R.H.S.=1.(1+3)4(1+1)(1+2)=1.44.2.3=16

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)=k(k+3)4(k+1)(k+2)…(i)11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)=k(k+3)4(k+1)(k+2)…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

[11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)]+1(k+1)(k+2)(k+3)[11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)]+1(k+1)(k+2)(k+3)

=k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3)=k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3) Using(i)Using(i)

=1(k+1)(k+2){k(k+3)4+1k+3}=1(k+1)(k+2){k(k+3)4+1k+3}

=1(k+1)(k+2){k(k+3)2+44(k+3)}=1(k+1)(k+2){k(k+3)2+44(k+3)}

=1(k+1)(k+2){k(k2+6k+9)+44(k+3)}=1(k+1)(k+2){k(k2+6k+9)+44(k+3)}

=1(k+1)(k+2){k3+6k2+9k+44(k+3)}=1(k+1)(k+2){k3+6k2+9k+44(k+3)}

=1(k+1)(k+2){k3+2k2+k+4k2+8k+44(k+3)}=1(k+1)(k+2){k3+2k2+k+4k2+8k+44(k+3)}

=1(k+1)(k+2){k(k2+2k+1)+4(k2+2k+1)4(k+3)}=1(k+1)(k+2){k(k2+2k+1)+4(k2+2k+1)4(k+3)}

=1(k+1)(k+2){k(k+1)2+4(k+1)24(k+3)}=1(k+1)(k+2){k(k+1)2+4(k+1)24(k+3)}

=(k+1)2(k+4)4(k+1)(k+2)(k+3)=(k+1)2(k+4)4(k+1)(k+2)(k+3)

=(k+1){(k+1)+3}4{(k+1)+1}{(k+1)+2}=(k+1){(k+1)+3}4{(k+1)+1}{(k+1)+2}

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

a+ar+ar2+…+arn-1=a(rn-1)r-1a+ar+ar2+…+arn-1=a(rn-1)r-1

P(n):a+ar+ar2+…+arn-1=a(rn-1)r-1P(n):a+ar+ar2+…+arn-1=a(rn-1)r-1

For n=1n=1,

L.H.S.=aL.H.S.=a

R.H.S.=a(r1-1)(r-1)=aR.H.S.=a(r1-1)(r-1)=a

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

a+ar+ar2+…+ark-1=a(rk-1)r-1…(i)a+ar+ar2+…+ark-1=a(rk-1)r-1…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

{a+ar+ar2+…+ark-1}+ar(k+1)-1{a+ar+ar2+…+ark-1}+ar(k+1)-1

=a(rk-1)r-1+ark=a(rk-1)r-1+ark Using(i)Using(i)

=a(rk-1)+ark(r-1)r-1=a(rk-1)+ark(r-1)r-1

=a(rk-1)+ark+1-arkr-1=a(rk-1)+ark+1-arkr-1

=ark-a+ark+1-arkr-1=ark-a+ark+1-arkr-1

=ark+1-ar-1=ark+1-ar-1

=a(rk+1-1)r-1=a(rk+1-1)r-1

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2

P(n):(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2P(n):(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2

L.H.S.=(1+31)=4L.H.S.=(1+31)=4

R.H.S.=(1+1)2=22=4R.H.S.=(1+1)2=22=4

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

(1+31)(1+54)(1+79)…(1+(2k+1)k2)=(k+1)2…(i)(1+31)(1+54)(1+79)…(1+(2k+1)k2)=(k+1)2…(i)

Consider

[(1+31)(1+54)(1+79)…(1+(2k+1)k2)]{1+{2(k+1)+1}(k+1)2}[(1+31)(1+54)(1+79)…(1+(2k+1)k2)]{1+{2(k+1)+1}(k+1)2}

=(k+1)2(1+2(k+1)+1(k+1)2)=(k+1)2(1+2(k+1)+1(k+1)2) Using(i)Using(i)

=(k+1)2[(k+1)2+2(k+1)+1(k+1)2]=(k+1)2[(k+1)2+2(k+1)+1(k+1)2]

=(k+1)2+2(k+1)+1=(k+1)2+2(k+1)+1

={(k+1)+1}2={(k+1)+1}2

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

(1+11)(1+12)(1+13)…(1+1n)=(n+1)(1+11)(1+12)(1+13)…(1+1n)=(n+1)

P(n):(1+11)(1+12)(1+13)…(1+1n)=(n+1)P(n):(1+11)(1+12)(1+13)…(1+1n)=(n+1)

For n=1n=1,

L.H.S.=(1+11)=2L.H.S.=(1+11)=2

R.H.S.=(1+1)=2R.H.S.=(1+1)=2

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

P(k):(1+11)(1+12)(1+13)…(1+1n)=(k+1)…(i)P(k):(1+11)(1+12)(1+13)…(1+1n)=(k+1)…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

[P(k):(1+11)(1+12)(1+13)…(1+1k)](1+1k+1)[P(k):(1+11)(1+12)(1+13)…(1+1k)](1+1k+1)

=(k+1)(1+1k+1)=(k+1)(1+1k+1) Using(i)Using(i)

=(k+1)[(k+1)+1(k+1)]=(k+1)[(k+1)+1(k+1)]

=(k+1)+1=(k+1)+1

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

12+32+52+…+(2n-1)2=n(2n-1)(2n+1)312+32+52+…+(2n-1)2=n(2n-1)(2n+1)3

P(n):12+32+52+…+(2n-1)2=n(2n-1)(2n+1)3P(n):12+32+52+…+(2n-1)2=n(2n-1)(2n+1)3

For n=1n=1,

L.H.S.=12=1L.H.S.=12=1

R.H.S.=1(2.1-1)(2.1+1)3=1.1.33=1R.H.S.=1(2.1-1)(2.1+1)3=1.1.33=1

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

P(k):12+32+52+…+(2n-1)2=k(2k-1)(2k+1)3…(i)P(k):12+32+52+…+(2n-1)2=k(2k-1)(2k+1)3…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

{12+32+52+…+(2k-1)2}+{2(k+1)-1}2{12+32+52+…+(2k-1)2}+{2(k+1)-1}2

=k(2k-1)(2k+1)3+(2k+2-1)2=k(2k-1)(2k+1)3+(2k+2-1)2 Using(i)Using(i)

=k(2k-1)(2k+1)3+(2k+1)2=k(2k-1)(2k+1)3+(2k+1)2

=2(2k-1)(2k+1)+3(2k+1)23=2(2k-1)(2k+1)+3(2k+1)23

=(2k+1){k(2k-1)+3(2k+1)}3=(2k+1){k(2k-1)+3(2k+1)}3

=(2k+1){2k2-k+6k+3}3=(2k+1){2k2-k+6k+3}3

=(2k+1){2k2+5k+3}3=(2k+1){2k2+5k+3}3

=(2k+1){2k2+2k+3k+3}3=(2k+1){2k2+2k+3k+3}3

=(2k+1){2k(k+1)+3(k+1)}3=(2k+1){2k(k+1)+3(k+1)}3

=(2k+1)(k+1)(2k+3)3=(2k+1)(k+1)(2k+3)3

=(k+1){2(k+1)-1}{2(k+1)+1}3=(k+1){2(k+1)-1}{2(k+1)+1}3

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)

P(n):11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)P(n):11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)

For n=1n=1,

L.H.S.=11.4=14L.H.S.=11.4=14

R.H.S.=13.1+1=14R.H.S.=13.1+1=14

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

P(k):11.4+14.7+17.10+…+1(3k-2)(3k+1)=k(3k+1)…(i)P(k):11.4+14.7+17.10+…+1(3k-2)(3k+1)=k(3k+1)…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

{11.4+14.7+17.10+…+1(3k-2)(3k+1)}+1{3(k+1)-2}{3(k+1)+1}{11.4+14.7+17.10+…+1(3k-2)(3k+1)}+1{3(k+1)-2}{3(k+1)+1}

=k3k+1+1(3k+1)(3k+4)=k3k+1+1(3k+1)(3k+4) Using(i)Using(i)

=1(3k+1){k+1(3k+4)}=1(3k+1){k+1(3k+4)}

=1(3k+1){k(3k+4)+1(3k+4)}=1(3k+1){k(3k+4)+1(3k+4)}

=1(3k+1){3k2+4k+1(3k+4)}=1(3k+1){3k2+4k+1(3k+4)}

=1(3k+1){3k2+3k+k+1(3k+4)}=1(3k+1){3k2+3k+k+1(3k+4)}

=(3k+1)(k+1)(3k+1)(3k+4)=(3k+1)(k+1)(3k+1)(3k+4)

=(k+1)3(k+1)+1=(k+1)3(k+1)+1

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

13.5+15.7+17.9+…+1(2n+1)(2n+3)=n2(2n+3)13.5+15.7+17.9+…+1(2n+1)(2n+3)=n2(2n+3)

L.H.S.=13.5L.H.S.=13.5

R.H.S.=13(2.1+3)=13.5R.H.S.=13(2.1+3)=13.5

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

P(k):13.5+15.7+17.9+…+1(2k+1)(2k+3)=k2(2k+3)…(i)P(k):13.5+15.7+17.9+…+1(2k+1)(2k+3)=k2(2k+3)…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

[13.5+15.7+17.9+…+1(2k+1)(2k+3)]+1{2(k+1)+1}{2(k+1)+3}[13.5+15.7+17.9+…+1(2k+1)(2k+3)]+1{2(k+1)+1}{2(k+1)+3}

=k3(2k+3)+1(2k+3)(2k+5)=k3(2k+3)+1(2k+3)(2k+5) Using(i)Using(i)

=1(2k+3)[k3+1(2k+5)]=1(2k+3)[k3+1(2k+5)]

=1(2k+3)[k(2k+5)+33(2k+5)]=1(2k+3)[k(2k+5)+33(2k+5)]

=1(2k+3)[2k2+5k+33(2k+5)]=1(2k+3)[2k2+5k+33(2k+5)]

=1(2k+3)[2k2+2k+3k+33(2k+5)]=1(2k+3)[2k2+2k+3k+33(2k+5)]

=1(2k+3)[2k(k+1)+3(k+1)3(2k+5)]=1(2k+3)[2k(k+1)+3(k+1)3(2k+5)]

=(k+1)(2k+3)3(2k+3)(2k+5)=(k+1)(2k+3)3(2k+3)(2k+5)

=(k+1)3{2(k+1)+3}=(k+1)3{2(k+1)+3}

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

1+2+3+…+n18(2n+1)21+2+3+…+n18(2n+1)2

P(n):1+2+3+…+n18(2n+1)2P(n):1+2+3+…+n18(2n+1)2

For n=1n=1,

118(2.1+1)2=98118(2.1+1)2=98

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,

1+2+3+…+k18(2k+1)2…(i)1+2+3+…+k18(2k+1)2…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.

Consider

1+2+3+…+k+(k+1)18(2k+1)2+(k+1)1+2+3+…+k+(k+1)18(2k+1)2+(k+1) Using(i)Using(i)

18{(2k+1)2+8(k+1)}18{(2k+1)2+8(k+1)}

18{4k2+4k+1+8k+8}18{4k2+4k+1+8k+8}

18{4k2+12k+9}18{4k2+12k+9}

18(2k+3)218(2k+3)2

18{2(k+1)+1}218{2(k+1)+1}2

Hence, 1+2+3+…+k+(k+1)18(2k+1)2+(k+1)1+2+3+…+k+(k+1)18(2k+1)2+(k+1)

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

n(n+1)(n+5)n(n+1)(n+5)is a multiple of 33.

P(n):n(n+1)(n+5)P(n):n(n+1)(n+5), which is a multiple of 33.

For n=1n=1,

1(1+1)(1+5)=121(1+1)(1+5)=12, which is a multiple of 33.

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some natural number kk, i.e.,

k(k+1)(k+5)k(k+1)(k+5) is a multiple of 33.

∴k(k+1)(k+5)=3m∴k(k+1)(k+5)=3m, where m∈Nm∈N …(i)

Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.

Consider

(k+1){(k+1)+1}{(k+1)+5}(k+1){(k+1)+1}{(k+1)+5}

=(k+1)(k+2){(k+1)+5}=(k+1)(k+2){(k+1)+5}

=(k+1)(k+2)(k+5)+(k+1)(k+2)=(k+1)(k+2)(k+5)+(k+1)(k+2)

={k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)={k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)

=3m+(k+1){2(k+5)+(k+2)}=3m+(k+1){2(k+5)+(k+2)}

=3m+(k+1){2k+10+k+2}=3m+(k+1){2k+10+k+2}

=3m+(k+1){3k+12}=3m+(k+1){3k+12}

=3m+3(k+1){k+4}=3m+3(k+1){k+4}

=3{m+(k+1)(k+4)}=3×q=3{m+(k+1)(k+4)}=3×q, where q={m+(k+1)(k+4)}q={m+(k+1)(k+4)} is some natural number.

Hence, (k+1){(k+1)+1}{(k+1)+5}(k+1){(k+1)+1}{(k+1)+5} is a multiple of 33.

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

102n-1+1102n-1+1 is divisible by 1111

P(n):102n-1+1P(n):102n-1+1 is divisible by 1111.

For n=1n=1,

P(1)=102n-1+1=11P(1)=102n-1+1=11 and P(1)P(1) is divisible by 1111.

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some natural number kk, i.e.,

i.e., 102k-1+1102k-1+1 is divisible by 1111.

∴102k-1+1=11m∴102k-1+1=11m, where m∈Nm∈N…(i)

Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.

Consider

102(k+1)-1+1102(k+1)-1+1

=102k+2-1+1=102k+2-1+1

=102k+1+1=102k+1+1

=102(102k-1+1-1)+1=102(102k-1+1-1)+1

=102(102k-1+1)-102+1=102(102k-1+1)-102+1

=102.11m-100+1=102.11m-100+1 Using(i)Using(i)

=100×11m-99=100×11m-99

=11(100m-9)=11(100m-9)

=11r=11r, where r=(100m-9)r=(100m-9) is some natural number

Therefore, 102(k+1)-1+1102(k+1)-1+1 is divisible by 1111.

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

x2n-y2nx2n-y2n is divisible by x+yx+y.

P(n):x2n-y2nP(n):x2n-y2n is divisible by x+yx+y.

For n=1n=1,

P(1)=x2×1-y2×1=x2-y2=(x+y)(x-y)P(1)=x2×1-y2×1=x2-y2=(x+y)(x-y), which is clearly divisible by x+yx+y.

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some natural number kk, i.e.,

x2k-y2kx2k-y2k is divisible by x+yx+y.

∴∴ Let x2k-y2k=m(x+y)x2k-y2k=m(x+y), where m∈Nm∈N…(i)

Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.

Consider

x2(k+1)-y2(k+1)x2(k+1)-y2(k+1)

=x2(x2k-y2k+y2k)-y2k.y2=x2(x2k-y2k+y2k)-y2k.y2

=x2{m(x+y)+y2k}-y2k.y2=x2{m(x+y)+y2k}-y2k.y2

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

32n+2-8n-932n+2-8n-9 is divisible by 88.

P(n):32n+2-8n-9P(n):32n+2-8n-9 is divisible by 88.

For n=1n=1,

P(n)=32×1+2-8×1-9=64P(n)=32×1+2-8×1-9=64, which is divisible by 88.

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some natural number kk, i.e.,

32k+2-8k-932k+2-8k-9 is divisible by 88.

∴32k+2-8k-9=8m∴32k+2-8k-9=8m; where m∈Nm∈N…(i)

Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.

Consider

32(k+1)+2-8(k+1)-932(k+1)+2-8(k+1)-9

32k+2.32-8k-8-932k+2.32-8k-8-9

=32(32k+2-8k-9+8k+9)-8k-17=32(32k+2-8k-9+8k+9)-8k-17

=32(32k+2-8k-9)+32(8k+9)-8k-17=32(32k+2-8k-9)+32(8k+9)-8k-17

=9.8m+9(8k+9)-8k-17=9.8m+9(8k+9)-8k-17

=9.8m+72k+81-8k-17=9.8m+72k+81-8k-17

=9.8m+64k+64=9.8m+64k+64

=8(9m+8k+8)=8(9m+8k+8)

=8r=8r, where r=(9m+8k+8)r=(9m+8k+8) is a natural number

Therefore, 32(k+1)+2-8(k+1)-932(k+1)+2-8(k+1)-9 is divisible by 88.

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

41n-14n41n-14n is a multiple of 2727.

P(n):41n-14nP(n):41n-14n is a multiple of 2727.

It can be observed that P(n)P(n) is true for n=1n=1

For n=1n=1,

P(1)=411−141=27P(1)=411−141=27, which is a multiple of 2727.

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some natural number kk, i.e.,

41k-14k41k-14k is a multiple of 2727.

∴41k-14k=27m,m∈N∴41k-14k=27m,m∈N …(i)

Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.

Consider

41k+1-14k+141k+1-14k+1

=41k.41-14k.14=41k.41-14k.14

=41(41k-14k+14k)-14k.14=41(41k-14k+14k)-14k.14

=41.27m+14k(41-14)=41.27m+14k(41-14)

=41.27m+27.14k=41.27m+27.14k

=27(41m-14k)=27(41m-14k)

=27×r=27×r, where r=(41m-14k)r=(41m-14k) is a natural number.

Therefore, 41k+1-14k+141k+1-14k+1 is a multiple of 2727.

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

(2n+7)(n+3)2(2n+7)(n+3)2

P(n):(2n+7)(n+3)2P(n):(2n+7)(n+3)2

For n=1n=1,

P(1)=2.1+7=9(1+3)2=16P(1)=2.1+7=9(1+3)2=16, which is true because 916916.

Therefore, P(n)P(n) is true for n=1n=1.

Let us assume that P(k)P(k) is true for some natural number kk, i.e.,

(2k+7)(k+3)2(2k+7)(k+3)2 …(i)

Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.

Consider

{2(k+1)+7}=(2k+7)+2{2(k+1)+7}=(2k+7)+2

∴{2(k+1)+7}=(2k+7)+2(k+3)2+2∴{2(k+1)+7}=(2k+7)+2(k+3)2+2 Using(i)Using(i)

2(k+1)+7k2+6k+9+22(k+1)+7k2+6k+9+2

2(k+1)+7k2+6k+112(k+1)+7k2+6k+11

Now, k2+6k+11k2+8k+16k2+6k+11k2+8k+16

∴2(k+1)+7(k+4)2∴2(k+1)+7(k+4)2

2(k+1)+7{(k+1)+3}22(k+1)+7{(k+1)+3}2

Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.

Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

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